Question:
Acceleration due to gravity at earth's surface is g $ms^{-2}$
Find the effective value of gravity at a height of 32 km from sea level $(R_e = 6400\, km)$
Acceleration due to gravity at earth's surface is g $ms^{-2}$
Find the effective value of gravity at a height of 32 km from sea level $(R_e = 6400\, km)$
Updated On: Jul 5, 2022
- $0.5\, g\, ms^{-2}$
- $0.99\, g\, ms^{-2}$
- $1.01\, g\, ms^{-2}$
- $0.90\, g\, ms^{-2}$
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The Correct Option is B
Solution and Explanation
For height h above the earth's surface
$\hspace20mm g' = g \bigg(1 - \frac{2h}{R_e}\bigg)=g \bigg(1 - \frac{64}{6400}\bigg)$
$\hspace70mm (\because R_e = 6400\, km)$
$\hspace25mm = g(1-0.01)$
$\hspace25mm = 0.99\, g\, ms^{-2}$
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Concepts Used:
Gravitation
In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.
Newton’s Law of Gravitation
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
- F ∝ (M1M2) . . . . (1)
- (F ∝ 1/r2) . . . . (2)
On combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2
The dimension formula of G is [M-1L3T-2].